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यदि x = -1 दोनों समीकरणों \(2x^2 + 3x + p = 0\) और \(qx^2 - qx + 4 = 0\) का एक उभयनिष्ठ मूल हो तो \(p + q\) का मान होगा

(A) 1

(B) -1

(C) 2

(D) -2

1 Answer

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Best answer

सही विकल्प है (B) -1

दिया है,

समीकरण \(2 x^{2}+3 x+p=0\) का मूल -1 है।

समीकरण \(2 x^{2}+3 x+p=0\) में x = -1 रखने पर

\(2(-1)^{2}+3(-1)+P =0\)

\(2 \times 1-3+P =0\)

\(2-3+P =0\)

\(-1+P=0\)

\(\therefore P =1\)

दूसरा समीकरण \(q x^{2}-q x+4=0\) है।

समीकरण \(q x^{2}-q x+4=0\) में x = -1 रखने पर

\(q(-1)^{2}-q(-1)+4 =0 \)

\(q+q+4 =0\)

\(2 q+4 =0\)

\(2q =-4 \)

\(\therefore q =\frac{-4}{2}=-2\)

इसलिए,

\(p+q\)

\(= 1+(-2)\)

\(= 1-2\)

\(= -1\)

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